Commit 8e2cb85

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feat: 136.single-number add Python3 implementation (azl397985856#91)

* feat: 136.single-number add Python3 implementation * update

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`@@ -36,7 +36,7 @@ Your algorithm should have a linear runtime complexity. Could you implement it w`
`36`	`36`
`37`	`37`	`## 代码`
`38`	`38`
`39`		`-* 语言支持:JS,C++`
	`39`	`+* 语言支持:JS,C++,Python`
`40`	`40`
`41`	`41`	`JavaScrip Code:`
`42`	`42`	```js
`@@ -109,6 +109,24 @@ public:`
`109`	`109`	`};`
`110`	`110`	```
`111`	`111`
	`112`	`+Python Code:`
	`113`	`+`
	`114`	+```python
	`115`	`+class Solution:`
	`116`	`+ def singleNumber(self, nums: List[int]) -> int:`
	`117`	`+ single_number = 0`
	`118`	`+ for num in nums:`
	`119`	`+ single_number ^= num`
	`120`	`+ return single_number`
	`121`	`+`
	`122`	`+# 另一种思路:集合是无序不重复的元素集`
	`123`	`+# 利用这一特性将列表转化成不包含重复元素的集合`
	`124`	`+# 分别对集合和列表进行求和,集合元素之和 x2 减去列表元素之和即为只出现了一次的元素`
	`125`	`+class Solution:`
	`126`	`+ def singleNumber(self, nums: List[int]) -> int:`
	`127`	`+ return 2 * sum(set(nums)) - sum(nums)`
	`128`	+```
	`129`	`+`
`112`	`130`	`## 延伸`
`113`	`131`
`114`	`132`	`有一个 n 个元素的数组,除了两个数只出现一次外,其余元素都出现两次,让你找出这两个只出现一次的数分别是几,要求时间复杂度为 O(n) 且再开辟的内存空间固定(与 n 无关)。`

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